Problem 92 - Calculus Early Transcendentals

Question

A focal chord of a conic section is a line through a focus joining two points of the curve. The latus rectum is the focal chord perpendicular to the major axis of the conic. Prove the following properties. The length of the latus rectum of the parabola \(y^{2}=4 p x\) or \(x^{2}=4 p y\) is \(4|p|\)

Step-by-Step Solution

Answer

Question: Prove that the length of the latus rectum for the parabolas \(y^2 = 4px\) and \(x^2 = 4py\) is \(4|p|\). Answer: Analyzing the two parabolas, we find their focus coordinates, vertex, and the equations of the axes. Then, we find the points on the parabolas where the latus rectum passes through and determine its length. For both parabolas, we find that the length of the latus rectum is \(4|p|\).

Step 1: Analyze the given parabolas and find the focus and vertex

For the first parabola, \(y^2 = 4px\), the vertex is at \((0,0)\) and the focus is situated on the x-axis at \((p,0)\). The axis of the parabola is the x-axis, or the line \(y=0\). For the second parabola, \(x^2 = 4py\), the vertex is at \((0,0)\), and the focus is situated on the y-axis at \((0,p)\). The axis of the parabola is the y-axis, or the line \(x=0\).

Step 2: Find the points on the parabolas where the latus rectum passes through

For the first parabola, the latus rectum is the horizontal line passing through the focus \((p,0)\). Let's call the two points on the parabola where it intersects the latus rectum \(A\) and \(B\). Since the latus rectum is perpendicular to the axis, it must be parallel to the y-axis. Therefore, both points A and B must have the same x-coordinate as the focus, which is \(p\). Now, let's substitute the x-coordinate of the points in the equation of the parabola, \(y^2 = 4px\): $$A(p, y_A): y_A^2 = 4p \cdot p \Rightarrow y_A = \pm 2p$$ $$B(p, y_B): y_B^2 = 4p \cdot p \Rightarrow y_B = \mp 2p$$ For the second parabola, the latus rectum is a vertical line passing through the focus \((0,p)\). Let's call the two points on the parabola where it intersects the latus rectum \(C\) and \(D\). Since the latus rectum is perpendicular to the axis, it must be parallel to the x-axis. Therefore, both points C and D must have the same y-coordinate as the focus, which is \(p\). Now, let's substitute the y-coordinate of the points in the equation of the parabola, \(x^2 = 4py\): $$C(x_C, p): x_C^2 = 4p \cdot p \Rightarrow x_C = \pm 2p$$ $$D(x_D, p): x_D^2 = 4p \cdot p \Rightarrow x_D = \mp 2p$$

Step 3: Find the length of the latus rectum for both parabolas

For the first parabola, the length of the latus rectum is the vertical distance between points A and B. Since \(y_A = 2p\) and \(y_B = -2p\), the length of the latus rectum for the parabola \(y^2 = 4px\) is: $$AB = |y_A - y_B| = |2p - (-2p)| = |4p|$$ For the second parabola, the length of the latus rectum is the horizontal distance between points C and D. Since \(x_C = 2p\) and \(x_D = -2p\), the length of the latus rectum for the parabola \(x^2 = 4py\) is: $$CD = |x_C - x_D| = |2p - (-2p)| = |4p|$$ Hence, the length of the latus rectum for both given parabolas is \(4|p|\).